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Algebra of Derivatives

In Δ ABC ...

Question

In $Δ$ $A B C$ the sides opposite to angles $A, B, C$ are denoted by $a, b, c$ respectively, then,
$(a^{2} - b^{2} - c^{2})$ $t a n A + (a^{2} - b^{2} + c^{2}) t a n B$

(a^{2} + b^{2} - c^{2}) t a n C

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(a^{2} + b^{2} + c^{2}) t a n C

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(a^{2} - b^{2} - c^{2}) t a n C

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Solution

The correct option is B $0$
$- (- a^{2} + b^{2} + c^{2}) tan A + (a^{2} - b^{2} + c^{2}) tan B$

$- (- a^{2} + b^{2} + c^{2}) tan A \times \frac{2 a b c}{2 a b c} + (a^{2} - b^{2} + c^{2}) tan B \times \frac{2 a b c}{2 a b c}$

now ,

$\frac{(b^{2} + c^{2} - a^{2})}{2 b c} = cos A$

$\frac{(a^{2} + c^{2} - b^{2})}{2 b c} = cos B$

&

$\frac{a}{sin A}$ = $\frac{b}{sin B}$ = $\frac{c}{sin C} = p$

$- \frac{(- a^{2} + b^{2} + c^{2})}{2 b c} \times \frac{sin A}{cos A} \times \frac{2 a b c}{a} + \frac{(a^{2} - b^{2} + c^{2})}{2 a c} \times \frac{sin B}{cos B} \times \frac{2 a b c}{b}$

$- \frac{2 a b c}{p} + \frac{2 a b c}{p} = 0$

Suggest Corrections

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